If $S=\{z\in\mathbb C:z\neq0\text{ and }\left|z+\frac{1}{z}\right|=a\}$ find $\min\limits_{z\in S}|z|$ 
If $a$ is a positive real number and $S=\{z\in \mathbb C : z\neq 0\ \text{ and } \vert z+\frac{1}{z}\vert = a\}$, Find $\min\limits_{z\in S}\vert z\vert$ and $\min\limits_{z\in S}\vert\frac{1}{z}\vert$

I first tried using the triangle inequality to get:
$$\vert z\vert^2-a\vert z\vert+1\ge 0$$
$$\vert z\vert\in\big(-\infty,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
since $\vert z\vert\ge 0$ and $a^2>a^2-4$,
$$\vert z\vert\in\big(0,\frac{a-\sqrt{a^2-4}}{2}\big)\cup\big(\frac{a+\sqrt{a^2-4}}{2},\infty\big)$$
So $\min\vert z \vert=0$. But the answer given is $\min\vert z\vert=\frac{\sqrt{a^2-4}-a}{2}$
 A: Let $z=re^{i\phi}$.$\,\,\,$  If $|z+z^{-1}|=a$, then 
$$r^2+r^{-2}+2\cos(2\phi)=a^2 \tag 1$$
We can write $(1)$ in a more convenient form as
$$\left(r+\frac1r\right)^2=a^2-4\sin^2(\phi) \tag 2$$
Note that $|a|\ge 2$ since the minimum of $r+\frac1r$ is $2$.
Now, differentiating both sides of $(2)$ with respect to $\phi$ yields
$$\left(2r-\frac{2}{r^3}\right)\frac{dr}{d\phi}=-4\sin(2\phi)$$
whereupon setting $\frac{dr}{d\phi}$ to $0$ reveals that $2\phi = \ell \pi$ for integer $\ell$.  
For $r\le 1$, the minimum for $r$ occurs when $2\phi =0$.  Substituting $2\phi=0$ into $(2)$ reveals
$$r_{min}+\frac1{r_{min}}=|a| \implies r_{min}=\frac{|a|-\sqrt{a^2-4}}{2}$$
A: Let $z=x+iy,\ \bar z = x-iy$.
If $\left|z+\frac 1 z \right| = a$, then 
$$z+\frac 1 z = a e^{it}$$ for some $t$.
This transforms into
$$z^2 - ae^{it}z+1=0.$$
Similarilry, we can get
$${\bar z}^2 - ae^{-it}\bar z+1=0.$$
Subtract these two equations; we get
$$z^2 -{\bar z}^2 -a(e^{it}z - e^{-it}\bar z)=0.$$
This is equivalent to
$$4ixy -2a\Im({e^{it}z})=4ixy-2a(x\sin t +y\cos t)=0.$$
For this equality to hold, we have one of two options:


*

*$x=0$, which implies that $t = \pi / 2$ since we don't want $y=0$,

*$y=0$, which implies that $t = 0 $ since we don't want $x=0$.


In the first case,
$$\left| iy+\frac {1}{iy}\right| = a$$
implies that 
$$y=\pm \frac{a\pm \sqrt{a^2-4}}{2}.$$ Since $a$ is positive, the minimum occurs at
$$y=\pm \frac{a- \sqrt{a^2-4}}{2}.$$
Therefore the minimum is 
$$ \frac{a- \sqrt{a^2-4}}{2}$$
Do a similar method for case $2$.
Hint for the second part: the minimum of $|1/z|$ is achieved at the maximum of $|z|$.
